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Which equation shows the vertex form of a quadratic equation with vertex (2, 5)?
A: f(x) = a(x – 2)^2 + 5
B: f(x) = a(x + 2) ^2+ 5
C: f(x) = a(x – 2)^2 – 5
D: f(x) = a(x + 2) ^2– 5


Sagot :

Answer:

Option A f(x) = a(x - 2)^2 + 5

Step-by-step explanation:

Vertex Form : y = a(x-h)^2 + k

In this case, we don't need the a-term because we are only looking for the vertex.

The h-term is how this parabola moved horizontally.

Since the x value is 2, that means that the graph moved 2 spaces towards the right. We plug it in to get:

y = (x - 2)^2

Now we need to find the k value. We see that the vertex is (2,5), and the y-value is 5. That means that the graph moved up to 5 units. Now we plug 5 into the k-value, and this is what we get:

y = (x - 2)^2 + 5

Walah! We're done with the equation, and as we can see, the answer is option A!

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